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Suggested solutions for Chapter 13 13 Problem 1 In the method for acetylating aromatic amines described in Chapter 8, 188 (the Lumière- Barbier method) the amine was dissolved in aqueous HCI. Using 1M amine, 4.6, and 1M HCI, -7, what will be the approximate equilibrium constant in the reaction? NH₂ + HX X The next step in the reaction is the addition of acetic anhydride and sodium acetate. What will happen to the sodium acetate (NaOAc) in the aqueous solution of HCI? Estimate the equilibrium constant for the reaction between NaOAc and HCI. Will there be enough acid left to keep the amine in solution? 0 + H NH₂ NH₃ N + HCI CI 0 0 Na It would be simpler not to add the sodium acetate, keeping the pH low, and thus definitely keep all the amine in solution. Why is it necessary to raise the pH for the second step of the reaction (the reaction with acetic anhydride)? Purpose of the problem Simple revision of carbonyl addition from Chapter 6 with a bit of substitution added. Suggested solution The answer to the first question is that the difference in values is very great (11.6 - remember that this means a factor of as it is a log scale) and there is enough acid to protonate all the amine. The equilibrium constant is enormous. In the second step, acetate ion about 5) and HCl -7) set up an equilibrium with acetic acid and chloride ion. The difference in values is still great and that equilibrium too will be right over on one side. However, the main point is that the pKₐ values of acetic acid and PhNH₃ are about the same so there will be about the same amount of each - the equilibrium constant will be about 1. Most of the amine will still be in solution as the cation but there will be enough free amine to react. If no sodium acetate were added, the amine would all be in solution as the cation and there would be essentially no free amine to act as a nucleophile. It would be simpler but no reaction would occur.